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Writing the wavefunction in modulus and dimensionless quantum phase gives a probability continuity equation and a Hamilton-Jacobi equation with a quantum potential:The phase equations hold locally away from wavefunction nodes. They follow by equating real and imaginary parts of the Time-dependent Schrodinger equation.

For unit mass and $\hslash =1$, this prepoint kernel tends to the identity and generates the Time-dependent Schrodinger equation. The square-root branch follows the damped Fresnel integral. Its normalized moments are $1$, $0$, and $i\Delta t$ at orders zero, one and two; these fix both the normalization and the kinetic sign.

Use the dimensionless quantum phase $S$, so that $\psi =R{e}^{iS}$ with $R\ge 0$. Work locally away from wavefunction nodes, where $R$ and $S$ are differentiable. Differentiating the wavefunction for substitution in the Time-dependent Schrodinger equation gives:Cancel $e^{iS}$ in the time-dependent equation and equate real and imaginary parts. The resulting Madelung equations areThe second equation is a Hamilton-Jacobi equation for the action $\hslash S$, with an additional quantum potential. To see the meaning of the first, multiply it by $2R$ and set $\rho =R^2$. It becomes the probability continuity equationThus the probability density is transported by the velocity field $\hslash \nabla S/m$. The Madelung equations are a local rewriting of the linear wave equation; their apparent nonlinearity comes from expressing a complex wavefunction in modulus and phase variables.

Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent isEach momentum integration is an ordinary Gaussian integral:Therefore the measure written in the question gives exactlyThe resulting configuration integral is $\int \mathcal{D}{q}_{\mathrm{raw},N}{e}^{-S_{E,N}}$. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses $d{p}_i/(2\pi )$ rather than $d{p}_i$. With that normalization the configuration measure isThis differs from the raw measure by $(2\pi {)}^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\dots ,q_N$ are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit $d{q}_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.

We now address the unheaded real-time continuation. Set $m=\hslash =1$. The normalized short-time Schrödinger kernel gives the final-slice recurrenceThe square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's $d{q}^{\prime }/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i{)}^{-1/2}$ fixes the identity limit and is included at every step.

Put $\eta =q^{\prime }-q$. The normalized oscillatory Gaussian has momentsTaylor-expand the smooth wavefunction and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order $(\Delta t{)}^2$. ThusSubtract the initial value, divide by $\Delta t$ and take the limit:This proves the Time-dependent Schrodinger equation by the Schrödinger equation from a short-time path integral argument. Nonuniform partitions give the same limit when their largest time step tends to zero.

The Born rule and probability current in one dimension giveMultiply the Time-dependent Schrodinger equation by ${\Psi }^{\ast }$, subtract its complex conjugate multiplied by $\Psi$, and rearrange to obtain the continuity equation ${\rho }_t+j_x=0$.

For the stated superposition, the interference terms in $j$ are real and cancel from its imaginary part. Henceindependent of $x$ and $t$.

Expand the state in the energy eigenstates of $H_0$ asSubstitution into the Time-dependent Schrodinger equation and projection onto $⟨k|$ gives the exact interaction picture equationsIn first-order time-dependent perturbation theory, put $c_j(t)={\delta }_{j0}$ on the right-hand side and integrate from $0$ to $t$:

For a uniform electric field, the electric-dipole interaction is $\Delta (t)=e{\mathcal{E}}_{0}z{e}^{-t/\tau }$ up to an irrelevant overall sign. The supplied hydrogen atom wavefunctions giveFor $t\gg \tau$, the time integral isThe long-time transition fraction is consequentlyBoth $|100⟩$ and $|200⟩$ have even parity, whereas $z$ is odd, so the electric-dipole selection rule gives $⟨200|z|100⟩=0$. Thus the requested fraction in $|200⟩$ is zero.

The Time-dependent Schrodinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.

The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

For the one-dimensional quantum harmonic oscillator,This is the Time-dependent Schrodinger equation with harmonic potential.

Expand the normalized initial wavefunction in the orthonormal energy basis:The Time-dependent Schrodinger equation then gives the unitary time evolution

The Time-dependent Schrodinger equation is $i\hslash {\partial }_t\psi =\hat{H}\psi$. Differentiate $⟨\hat{A}⟩=⟨\psi |\hat{A}|\psi ⟩$, use the equation and its adjoint, and include explicit time dependence of the operator. This gives Ehrenfest theorem

Insert the stated expansion into the Time-dependent Schrodinger equationWhen the derivative acts on each free phase, its energy factor is cancelled by the action of $H_0$ on the corresponding eigenstate. The remaining terms obeyDivision by $i\hslash$ gives

Expand a smooth wavefunction and potential in the normalized short-time Schrödinger kernel. Its odd moment vanishes and its second moment is $i\Delta t$, so one step changes the wavefunction by $i\Delta t{\psi }^{\prime \prime }/2-i\Delta tV\psi +O(\Delta t^2)$. Division by the time step gives the Time-dependent Schrodinger equation.